How do you solve #Log_a x - Log_a 2 = Log_a 9 + Log_a 8#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer ali ergin Mar 6, 2016 #x=144# Explanation: #log_a( x/2)=log_a3^2+log_a2^3# #log_a( x/2)=log_a(3^2*2^3)# #x/2=3^2*2^3# #x=3^2*2^4# #x=9*16# #x=144# Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 1167 views around the world You can reuse this answer Creative Commons License